First of all, I define accumulation point as: Let $a$ be an accumulation point of $A$. $\forall \ \epsilon > 0$, $B_\epsilon(a) \setminus \{a\} $ contains an element of $A$.
T/F question: If $a$ is an accumulation point of $A \subset \mathbb{R}$, then there exists a sequence $(a_n)$ in $A$ converging to $a$.
I'm tempted to say true by using this proof:
For each $n \in \mathbb{N}$, construct $B_\epsilon(a)$ such that $\epsilon = 1/n$. Then pick a point out of the ball such that $a_n \neq a$. Repeating this procedure yields a sequence and this sequence converges to $a$.
Two questions:
(1) Am I right
(2) Is it necessary that $A$ is a subset of $\mathbb{R}$? Why or why not?